Question

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
${}^{\prime }{\alpha }^{\prime }$ is the degree of dissociation of $PC{l}_5$ at equilibrium pressure ${}^{\prime }{P}^{\prime }$. Which among the following is the correct expression for degree of dissociation of ${}^{\prime }{\alpha }^{\prime }$ ?

• A. $\alpha =\sqrt{\frac{K_P}{P+K_P}}$
• B. $\alpha =\sqrt{\frac{P+K_P}{K_P}}$
• C. $\alpha =\sqrt{K_{P}P}$
• D. $\alpha =\sqrt{P/K_P}$

Answer 1

The correct option is A $\alpha =\sqrt{\frac{K_P}{P+K_P}}$

$PC{l}_5\to PC{l}_3+C{l}_2$

1 0 0 (Conc. at starting)

$1-\alpha$ $\alpha$ $\alpha$ (At Eqilibrium)

Then total no. of moles =$(1-\alpha )+\alpha +\alpha =1+\alpha$

Let total pressure be P.

Partial pressure of:-

$PC{l}_5=\left[\right(1-\alpha \left)/\right(1+\alpha \left)\right]\times P$

$PC{l}_3=\left[\right(\alpha \left)/\right(1+\alpha \left)\right]\times P$

$C{l}_2=\left[\right(\alpha \left)/\right(1+\alpha \left)\right]\times P$

$K\left(P\right)=\frac{\left[\right(\alpha \left)/\right(1+\alpha \left)\right]\times P\times \left[\right(\alpha )}{(1+\alpha )]\times P]/\left[\right(1-\alpha \left)/\right(1+\alpha \left)\right]\times P}$

$=\alpha \times \alpha \times P/(1-{\alpha }^2)$

$\alpha =\sqrt{\frac{K_P}{P+K_P}}$A is correct answer